Unit 2
Circuit Analysis Techniques
In this unit we apply our knowledge of KVL, KCL and Ohm’s Law to develop further
techniques for circuit analysis. The material is based on Chapter 4 of the text and
that chapter should be read in conjunction with the notes. For the most part we will
study the ‘new’ techniques by way of examples.
Again, see the definitions in Table 4.1 of the text. In particular we note the
following which will be directly relevant in this Unit:
(1) An essential node is a point in a circuit where three or more circuit elements
join.
(2) An essential branch is a path that connects two essential nodes without passing
through an essential node.
In solving the circuits of this Unit – that is, in finding any unknown parameters
(voltages, currents or resistances) – of course, we must analyze the circuit to obtain
the same number of independent equations as there are unknowns. We have already
seen that if there are n nodes in a circuit, we may obtain n − 1 equations by using
KCL. If there are b unknown currents in total, then clearly b equations are required
so that b − (n − 1) equations must be found by some other means – KVL is a good
candidate. Remember that the b equations must be independent. Deciding on which
techniques to use in developing the minimum number of equations required takes
PRACTICE.
2.1
The Node-Voltage Method
In circuit analysis it is often useful to consider one node in a circuit to be a reference
node relative to which the voltages at the other (non-reference) nodes are measured.
It also usually a good idea to choose the reference node as that node connected to
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the most branches. If there are ne essential nodes in a circuit, the circuit may be
described by ne − 1 equations.
A node voltage is defined as the voltage rise from a reference node to a nonreference node. In the circuit shown below there are only 3 essential nodes. This is
emphasized by labelling one node twice – eg., electrically, the two cs are the same
point. It may be noted that the following node-voltage technique is really a result of
KCL. We will now use the node-voltage method to determine the (i) voltages across
the 1 Ω and 12 Ω resistors, (ii) v as shown, and (iii) subsequently, we’ll use the results
to find the current through the 12 Ω resistor.
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Observation: Note that since there are 3 essential nodes, we may describe the circuit
with 2 (i.e. one less than the number of essential nodes) node-voltage equations.
Step 1: Choose a reference node (indicated by the solid arrow) – notice it’s the node
with the most branches.
Step 2: Label the remaining essential nodes with their voltages with respect to the
reference. Here we have used v1 and v2 .
Step 3: Use KCL at each of the nodes in Step 3. In doing this the text’s convention
is to sum the currents leaving each of these nodes to zero. Here, we get
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Step 4: Use the node-voltages to determine any required branch currents and powers.
(NOTE: An example involving dependent sources and supernodes – i.e. nodes between
which a voltage source exists – will be completed in a Tutorial.)
It may be noted that if a voltage source, vs , is connected directly between two essential
nodes, then it constrains the voltage difference between these nodes to have a value vs .
For example, in the diagram below, where the lower node is taken to be the reference,
we know that v1 = vs without having to apply KCL at node 1 – this reduces the
number of unknown voltages and simplifies the analysis.
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2.2
Mesh-Current Method
Recall that a mesh is a circuit loop that does not enclose any other loops. In the
circuit shown below there are 3 meshes. A mesh current may be thought of as a
current that exists in the perimeter of a mesh. This need not be an actual current,
but as we will see, actual currents may be determined from the mesh currents as
we shall see in the following example. The mesh-current method of circuit analysis
actually makes use of KVL for individual meshes.
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Step 1: Assign a mesh current to each of the meshes as indicated by the arrowed
curves labelled i1 , i2 , and i3 .
Step 2: Apply KVL to each mesh to obtain a system of equations involving the mesh
currents. Solve these for the mesh currents.
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Step 3: Obtain any necessary branch currents from the mesh currents. This allows
any voltages and powers of interest to be calculated. Here we illustrate by finding (a)
the power delivered by the 80 V source and (b) the power dissipated in the 8 Ω and
26 Ω resistors. (It is also easy to check that “power delivered” = “power absorbed”.)
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Extra notes on mesh currents:
(1) If there is a dependent source in the circuit, the mesh equations will have to be
supplemented by an equation involving that source.
Illustration:
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(2) When a branch contains a current source, the voltage across it is not known. Then
we may “ignore” that branch and create a supermesh which avoids it. The supermesh
must use the same mesh currents as originally chosen and a separate equation relating
the previously “ignored” source to these currents will be required also.
Illustration:
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2.3
Source Transformations and Equivalent Circuits
Source Transformations
It is possible to replace a voltage source in series with a resistor with a current
source in parallel with the same resistor (or vice versa) so that the voltage vab across
a load, say RL , remains the same. Consider the following in which we require vab to
be the same in each case:
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In the first circuit,
vab =
. . . (1)
while in the second,
vab =
. . . (2)
From (1) and (2), it is thus required that
This means that
is =
vs
. . . (3).
R
As far as the behaviour at the terminal a and b is concerned, both circuits are equivalent if (3) holds.
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Thévenin Equivalent Circuits
Consider a resistive network possibly containing both dependent and independent
sources. We wish to reduce the circuit to a single equivalent independent voltage and
resistance such that the current and voltage associated with a load connected across
a and b will be the same in the simplified circuit as it was in the original network.
Illustration:
When this is the case, the voltage and resistance are referred to as the Thévenin
voltage and resistance, symbolized as VTh and RTh , respectively. This equivalence
must hold for all possible values of load resistance RL . It must therefore hold in the
two extreme cases: (i) RL → ∞ (i.e. an open circuit) and (ii) RL → 0 (i.e. a short
circuit). In the first of these cases,
vab = VTh = voc
. . . (4)
where voc is the open-circuit voltage. Since, by definition, the Thévenin voltage is the
same for any load, equation (4) indicates we may get it by simply open-circuiting the
load. In the second case, the short-circuit current isc must be given by
isc =
VTh
RTh
. . . (5)
again since both VTh and RTh are required to be unchanged for any load. From (5)
RTh =
VTh
isc
. . . (6)
and equations (4) and (6) thus define the Thévenin equivalent circuit.
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Norton Equivalent Circuit
Before elaborating on Thévenin’s equivalent circuits we note the following source
transformation:
Illustration:
From our discussion on source transformations at the beginning of this subsection,
we know that the above circuits are equivalent as long as
is =
In this case the circuit on the right is referred to as the Norton equivalent circuit.
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Preliminary Example on Thévenin/Norton Equivalent Circuits
Find the Thévenin and Norton equivalent circuits for
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Thévenin Equivalent Circuit:
Using source transformation, the circuit may be represented as
The Thévenin voltage and resistance may calculated simply as follows:
Norton Equivalent Circuit:
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More on Thévenin Equivalent Circuits
Depending on the type of circuit, reduction to its Thévenin equivalent may involve a
variety of techniques:
(1) In general, voc , isc , and/or RTh may be determined using the results of nodevoltage and mesh-current analysis.
(2) As above, sometimes use of equivalent sources is helpful.
(3) If the circuit contains only independent sources, RTh may be obtained easily by
(i) short-circuiting the voltage supplies and (ii) open-circuiting the current supplies.
This works only because RTh does not depend on the values of the independent voltage
and current supplies in the original circuit. Consider this for the above example:
(4) If the circuit contains dependent sources, RTh may be determined by (i) shortcircuiting the independent voltage supplies, (ii) open-circuiting the independent current supplies and (iii) connecting a “test” source across the terminals and determining
the equivalent resistance of the revised circuit from its response to the test source.
Consider Drill Exercise 4.20, p. 119 of the text:
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2.4
Preliminary Look at Maximum Power Transfer
Later we shall be concerned with maximum power being delivered from a circuit which
has elements other than resistors to a load which is not purely resistive. Here, however,
we review the idea of maximum power transfer from a purely resistive network to a
purely resistive load. The purely resistive network may be represented by its Thévenin
equivalent and is depicted below as being attached to a load resistance RL . We wish
to establish the value of RL that permits maximum power transfer from the network
to the RL .
Illustration:
We begin by noting that power delivered to the load is a function of RL as given by
p = i2 RL =
=
. . . (1)
Since we wish to determine the value of RL which maximizes p, we may take the first
derivative of p in (1) with respect to RL , set the result to 0 and solve for RL :
dp
=
dRL
=
12
. . . (2)
Clearly, RL = RTH produces an extreme value. The second derivative of p with respect
to RL is negative (SHOW THIS) for this value. Thus,
RL = RTH
. . . (3)
is the value of RL which maximizes the power delivered to the load. From (1), this
power is given by
pmax =
. . . (4)
Example: Similar to Problem 4.79, page 139 of text.
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2.5
Superposition
The principle of superposition states that whenever a linear system is driven by more
than one independent source of energy, the total response of the system equals the
sum of the responses from each of the sources acting individually. The application of
superposition to the problems encountered thus far, in which all independent sources
are dc, does not offer an improvement over the techniques already used. However, for
illustrative purposes we’ll consider superposition briefly.
In applying superposition: (i) to find the effects of independent voltage sources
(one at a time) the independent current sources are deactivated by open-circuiting
them; (ii) to find the effects of independent current sources (one at a time) the independent voltage sources are deactivated by short-circuiting them; (iii)dependent
sources are never deactivated.
Example: Consider the following circuit. We wish to find the voltage v as shown.
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